Find the probability that a patient is able to be served immediately on arrival Somewhere in a rural region in Africa, patients arrive a mobile vaccination clinic. The can which is staffed by one nurse faces random demand at a mean rate of $15$ per hour. The time required for patients to get their vaccination has the negative exponential distribution with mean 5 minutes.
$a)$ Find the probability that a patient is able to be served immediately on arrival
$b)$ Calculate the most probable number of patients to be treated if in both cases there is one nurse.
$c)$ Determine the probability that there are 3 patients in the system
I have begun to answer $a)$ Not sure if I am on the right lines or not.
Average 10 minutes between arrivals $\lambda = \frac{1}{15}$
Average length of service is $\mu= \frac{1}{3}$
so P(not in use) $=\frac{2}{3}$
$b)$ $L=\frac{p}{1-p}= \frac{1}{2}$
Not too sure about C
 A: This is a $M/M/1$ queue with $\lambda=4$ and $\mu=5$.Let $X_n$ be the patients in the system at time $n$, then $\{X_n:n\geqslant0\}$ is a Markov chain on $\mathbb Z_{\geqslant0}$ with transition probabilities
$$
P_{ij} =\begin{cases}
\lambda,& j=i+1\\
\mu,& j=i-1\\
0,& \text{otherwise}.
\end{cases}
$$
Since $\rho:=\lambda/\mu=4/5<1$, the chain is ergodic and has a unique stationary distribution $\pi$ satisfying the detailed balance equations $\pi_iP_{ij} = \pi_jP_{ji}$, or $$\lambda\pi_{n-1} = \mu\pi_n,\quad n\geqslant 1. $$
This yields $\pi_n=\rho\pi_{n-1}$, and by recursion $\pi_n=\rho^n\pi_0$. From $\sum_{n=0}^\infty \pi_n=1$ we find that $\pi_0 = 1-\rho$ and hence $\pi_n=\rho^n(1-\rho)$. Substituting $\lambda=4$ and $\mu=5$, we have $\pi_n=\left(\frac45\right)^n\frac15$.
The probability that a patient is served immediately on arrival is the limiting probability that the system is empty, i.e. $\pi_0=\frac15$.
The question in b) is stated a bit oddly, but it seems to be asking for $i^* := \mathrm\arg\max_i \pi_i$. We can compute the limiting average number of customers in the system by
$$
L = \sum_{n=1}^\infty n\pi_n = \sum_{n=1}^\infty n\left(\frac45\right)^n\frac15 = 4,
$$
and since this is an integer, $i^*=4$.
The limiting probability that there are three patients in the system is given by
$$
\pi_3 = \left(\frac45\right)^3\frac15 =\frac{64}{125}\approx0.1024.
$$
